Why can't a set have two elements of the same value?

Question

Suppose I have two sets, $A$ and $B$:

$$A = \{1, 2, 3, 4, 5\} \\ B = \{1, 1, 2, 3, 4\}$$

Set $A$ is valid, but set $B$ isn't because not all of its elements are unique. My question is, why can't sets contain duplicate elements?

Answer 1

The short, perhaps unsatisfying answer is, because that is how they are defined. The long answer is that, in most cases, that is what is useful.

For other cases, there is also a theory built around multisets, which are like sets except they allow multiplicity.

Answer 2

I'd say that $B$ is valid and equal to $\{1, 2, 3, 4\}$.

The notation $B = \{1, 1, 2, 3, 4\}$ gives $B$ by listing its elements:

$1 \in B$

$1 \in B$

$2 \in B$

$3 \in B$

$4 \in B$

Clearly saying twice that $1 \in B$ is harmless.

This is the axiom of extensionality: two sets are equal iff they have the same elements.

Answer 3

Do not think of 1 and 1 as "two elements of the same value". They are the same element really. And an element is either a member of a set or it is not.

Answer 4

Short Answer:

The axioms of the Set Theory do not allow us to distinguish between two sets like $A = \{1,2\}$ and $B = \{1,1,2\}$. And every sentence valid about sets should be derived in some way from the axioms.

Explained:

The logic of the set theory is extensional, that means that doesn't matter the nature of a set, just its extension. The set $A = \{1,1,2,3,4\}$ could be considered different from $B = \{1,2,3,4\}$ in intension, but they are not different in extension, since $1 = 1$, both sets have the same elements. Even if one instance of the number $1$ precedes the other, the axioms of the set theory do not allow to distinguish between two sets using the order of its elements, because the notion of order is not defined by the axioms of the theory. To do that, the ordered pair is defined as follows: $$\left(a,b\right) \equiv \{\{a\},\{a,b\}\}$$ Thus $\left(a,b\right) \neq \left(b,a\right)$ since $\{\{a\},\{a,b\}\} \neq \{\{b\},\{b,a\}\}$. Thus, the set $A$ can't be distinguished fom B using its order. Moreover, the cardinality of two sets is never used to prove that two sets are different.

What about the ordered pair $\left(1,1\right) \equiv \{\{1\},\{1,1\}\} $? Since $1 = 1$ we have $\{1,1\} \equiv \{1\}$, this pair is a well known structure called singlenton, obtained as a consequence of the axiom of pairing and (as you can see) the axiom of extension. Moreover, since $\{1\}$ is present in $\left(1,1\right)$ two times, $\left(1,1\right) \equiv \{\{1\}\}$ another singlenton. Which is not a problem for the Analysis in $\mathbb{R}^2$, because this pair is different from $\{1\}$, a subset of $\mathbb{R}$, and different from each other pair in $\mathbb{R}^2$. The moral of this story is that you can't distinguish between the first and the second element of $\left(1,1\right)$, and it's ok.

Although the symbol "$\equiv$" is used to introduce syntactic abbreviations instead of semantic equality denoted by "$=$", this difference is just important in the study of Formal Languages, but not really important for this case, since the axiom of extension is what allows us to consider $\{1,1\} \equiv \{1\}$ as $\{1,1\} = \{1\}$. Just read the Section 3 of the book Naive Set Theory of Paul R. Halmos, what introduces the term singlenton in that way, and that's how is used by the specialized literature.

Other reference to understand the difference between intension and extension.

https://plato.stanford.edu/entries/logic-intensional/

Answer 5

Informally, the "set" of members of your household should have a well defined size, and the number of nicknames a person has is unimportant to that set.

It's just agreed by convention that no matter how many labels for the same thing we try to include in a set, the set only contains the things themselves. You refer to 1 twice, but that's just repeating the name of the thing, and by definition, the set only contains the things themselves.

Answer 6

If we had a 'set' $U$ defined to be $\{1,2,3\}$, could we have another set $V = \{1,1,2,3\}$ where $U\neq V$?

Yes. We can do this, but using this notation makes it ambiguous and nasty.

We need to be careful about what we mean when we say $=$. Let's let $=_S$ denote an equivalence relation on sets, and $=_n$ be the usual equivalence relation of numbers.

It is perfectly consistent to talk about $U,V$, but if we write it down like this we've labeled everything using the weaker relation $=_n$, making it a big mess. I'll show you what I mean:

If $a,b$ are two distinct elements in $V$ where $a=_n 1$ and $b=_n 1$, then $a=_nb$. Furthermore, if we choose the $c \in U$ where $c=_n1$, then $a =_n b =_n c$.

We want $U\neq _S V$, so we must have that $\{a\} \neq_S \{b\}$. We have no idea whether or not $\{c\}=_S \{a\}$ or $\{b\}$. In fact, it might not even be equal (in the set sense) to either of them.

With this knowlege we can re-label the $1$s in our sets: $U=\{1_c, 2, 3\}$, and $V=\{1_a, 1_b, 2, 3\}$. But we still don't have any idea whether or not $1_c$ as a set element is the same as $1_a$ or $1_b$. The same can be said for $U$'s $2$ and $3$ vs $V$'s $2$ and $3$.

Under this notation we have no idea how $=_S$ works comparing elements across sets.

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Here's a better way to do it. Instead of worrying about the set relation, define U and V using an index.

To start, clarify that every time you write down $a_i$, $i\in \mathbb{N}$, you are talking about the exact same object $a_i$, in every context. This forces the following property: $\{a_i\} =_S \{a_i\}$.

Now define $a_1 =_n 1\in \mathbb{N}; a_2 =_n 1 \in \mathbb{N}; a_3 =_n 2 \in \mathbb{N}; a_4 =_n 3 \in \mathbb{N}$, and let $U=\{a_1, a_2, a_3, a_4\}, V=\{a_2, a_3, a_4\}$

What we have really done now is make a function from the set $\{1,2,3,4\}\subset \mathbb{N}$ into the set $\{1,2,3\}\subset\mathbb{N}$, and defined $U, V$ in terms of that function. Under this description the $=_S$ relation is well-defined.

Answer 7

To say $2$ sets are equal, show that each set is contained in the other. $\{1,1,2,3,4\} = \{1,2,3,4\}$ since any element on the left can be found on the right and vice-versa.

Answer 8

The problem here is one of equating the definition of a set with its representation. The fact that you can kind of represent a set on paper doesn't mean that that representation is the set. So, for example, I can talk about the set that only contains the number we call one, and I can make this concrete by writing it thus: $\{1\}$. If I then wrote $\{1, 2\}$ and said that this is the same set then you would correctly object; but if I wrote $\{1, 1\}$ and said that this was the same set then you should not object because that depiction satisfies the definition.

It's like having the definition of $\pi$ as the ratio of a circle's circumference to its diameter and then writing a number like 3.14159 and saying that the number is $\pi$ - $\pi$ is the definition and the number is what we write to represent $\pi$, and that representation can be imprecise or deficient.

Answer 9

It stated that members of a set should have a common character so that we can easily identify the non-members of a given set. Supposing we have: S={ 2,4,6,8,10...} The comm characteristics that exist among the members of set S is that. We always get the same number when we subtract on member from the other member at the immediate next. I. e 4-2=2 6-4= 2 8-6=2. The two(2)s that appears is a unique feature the sets have.

But if say A= { 1,1,2, 1,3 4...} It will be difficult to get the unique features the exist between the of the set. Hence, set A may be said to be not well defined set.